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Deepika bought 7 packets of milk. Each packet contained 0.45 L of milk. How much milk did she buy altogether? \[\]
A.$0.315L$\[\]
B.$3.15L$\[\]
C.$31.5L$\[\]
D.$315L$\[\]

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Last updated date: 26th Jul 2024
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Answer
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Hint: We use the unitary method for direct variation and multiply the amount of milk in one packet 0.45L to the number of packets Deepika bought. We use the usual method of decimal multiplication and multiply 7 and 45 and put the decimal point at the right place in the product. \[\]

Complete step-by-step answer:
We know that the unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value. There are two types of two types of unitary method one is direct variation and other is indirect variation. \[\]
When one quality $a$ increases with another quantity $b$ and also $a$ decreases with $b$ we say the quantities $a$ and $b$ are in direct variation. We divide the increasing quantity $a$ by $b$ to obtain the value of single unit and then multiply to find the required value.\[\]
We are given in the question that Deepika bought 7 packets of milk. Each packet contained 0.45 L of milk. Here we are already given the value of a single unit that is the volume of single packet milk 0.45L. So just need to multiply with the number of packets to get the result. So she will total amount milk is 7 times the amount of milk in one packet that is $7\times 0.45$L. \[\]
We know how to multiply decimals. We forget the decimal point and just multiply$7\times 45=315$. We see the total number of digits after decimal is in the multiplication of $7\times 0.45$ as 2. So we put the decimal point in 315 so that 2 digits will remain after the decimal point. So the answer is 3.15L. Hence the correct option is B. \[\]

So, the correct answer is “Option B”.

Note: We see that in direct variation the fraction $\dfrac{a}{b}$ always remains constant. The problems of weight, length and volume are in direct variation. We also note that we do not count 0 as a digit after decimal point unless a non-zero digit appears after that, for example in $1.0450$ we count only $0,4,5$and have the number of digits as 3.